Figure 13.1 shows a body of arbitrary shape balanced by a single
force. The origin of the coordinate system is defined such that it coincides
with the center of gravity of the object, which is the point upon which
the balancing force acts. An object that is supported at its center of gravity
will be in static equilibrium, independent of the orientation of the object.
If the body is in equilibrium, the net force acting on it must be zero. Figure
13.1 shows that

Since the body is in equilibrium

and therefore

In obtaining this result we have assumed that the gravitational
acceleration is the same for every point of the body. The net torque
acting on the body is given by

Since the body is in static equilibrium

and therefore

This shows that rcm = 0 or rcm is parallel to g.
We conclude that for a body to be in equilibrium, its center of mass must
coincide with its center of gravity.

Sample Problem 13-1

A uniform beam of length L whose mass is m, rest with its ends on two
digital scales (see Figure 13.2). A block whose mass is M rests on the beam,
its center one-fourth away from the beam's left end. What do the scales read
?

Figure 13.2. Sample problem 13-1.

For the system to be in equilibrium, the net force and net torque
must be zero. Figure 13.2 shows that

Here we have replaced the force acting on the beam with a single force
acting on its center of gravity. The net torque of the system, with respect to
the left scale, is

This shows immediately that

From the equation of the net force we obtain

Sample Problem 13-3

A ladder with length L and mass m rests against a wall. Its upper
end is a distance h above the ground (see Figure 13.3). The center of gravity
of the ladder is one-third of the way up the ladder. A firefighter with mass M
climbs halfway up the ladder. Assume that the wall, but not the ground, is
frictionless. What is the force exerted on the ladder by the wall and by the
ground ?

The wall exerts a horizontal force FW on the ladder (the normal
force); it exerts no vertical force. The ground exerts a force Fg
on the ladder with a horizontal component Fgx and a vertical
component Fgy. If these two components were not present, the system
would not be in equilibrium. The net force in the x and y directions is given
by

and

The net torque, with respect to O (which is the contact point between
the ladder and the ground), is given by

Figure 13.3. Sample Problem 13-3

This immediately shows that

We can now calculate the force Fg:

and

We observe that Fgx depends on the position of the
firefighter. Suppose that the firefighter is a distance f L up the ladder. In
this case Fgx is given by

If the coefficient of static friction between the ladder and the ground
is us, than the maximum distance the firefighter can climb is
reached when

Two bricks of length L and mass m are stacked. Using conditions of
static equilibrium we can determine the maximum overhang of the top brick (see
Figure 13.4).

The two forces acting on the top brick are the gravitational force
Fg and the normal force N, exerted by the bottom brick on the top
brick. Both forces are directed along the y-axis. Since the system is in
equilibrium, the net force acting along the y-axis must be zero. We conclude
that

Figure 13.4. Two stacked bricks.

If the top block is on the verge of falling down, it will rotate
around O. The torque exerted by the two external forces with respect to O can
be easily calculated (see Figure 13.5). The gravitational force Fg
acting on the whole block is replaced by a single force with magnitude m g
acting on the center of mass of the top block. The normal force N acting on
the whole contact area between the top and the bottom block is replaced by a
single force N acting on a point a distance d away from the rotation axis O.
The torque of the normal force and the gravitational force with respect to O is
given by

The net torque acting on the top brick is given by

If the system is in equilibrium, then the net torque acting on the top
brick with respect to O must be zero. This implies that

Figure 13.5. Forces acting on top brick.

or

This equation shows that the system can never be in equilibrium if a
> L/2 (since d < 0 in that case). The system will be on the verge of
losing equilibrium if a = L/2. In this case, d = 0. We conclude that the
system can not be in equilibrium if the center of mass of the top brick is
located to the right of the edge of the bottom brick. the system will be on
the verge of losing equilibrium if the center of mass of the top brick is
located right over the edge of the bottom brick. Finally, if the center of
mass of the top brick is located to the left of the edge of the bottom brick,
the system will be in equilibrium.